Realistic Math Problems Help 6th-graders Solve Real-Life Questions ... - Free Printable

Let's go through each of the word problems one by one, solve them step-by-step, and verify the given answers.

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Problem 1:

"I'm thinking of two numbers, 12 and another number. 12 and my other number have a greatest common factor of 6 and their least common multiple is 36. What's the other number I'm thinking of?"

Given:
- One number = 12
- GCF (Greatest Common Factor) of 12 and the unknown number = 6
- LCM (Least Common Multiple) of 12 and the unknown number = 36

We use the formula:
> GCF × LCM = Product of the two numbers

So:
> $ 6 \times 36 = 12 \times x $

Calculate:
> $ 216 = 12x $

Divide both sides by 12:
> $ x = 216 ÷ 12 = 18 $

✔ So the other number is 18

Check:
- GCF of 12 and 18 = 6 ✔
- LCM of 12 and 18 = 36 ✔

✔️ Answer: 18 — Correct.

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Problem 2:

"Tom had a platter of chocolate wafers. He ate 5 of them and then gave his brother 3, he then handed them to his ball team of 8 members. The first player to arrive took 1, the second player took 3, the third player took 5 and so on. When the last player took his, the platter was empty. How many chocolate wafers did Tom start with?"

Step-by-step:
- Tom ate: 5
- Gave to brother: 3
- So far: 5 + 3 = 8 wafers used
- Then 8 players took wafers in an odd-numbered pattern: 1, 3, 5, 7, 9, 11, 13, 15

This is an arithmetic sequence:
- First term $ a = 1 $
- Common difference $ d = 2 $
- Number of terms $ n = 8 $

Sum of an arithmetic sequence:
> $ S_n = \frac{n}{2} (2a + (n-1)d) $

Plug in:
> $ S_8 = \frac{8}{2} (2(1) + (8-1)(2)) = 4 (2 + 14) = 4 × 16 = 64 $

So the 8 players took 64 wafers

Total wafers:
> Tom’s 5 + brother’s 3 + players’ 64 = 72

✔ Answer: 72 — Correct.

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Problem 3:

"Jasmine has 50 marbles in a bag. 20% of the marbles are blue. How many marbles are blue?"

20% of 50:
> $ 0.20 × 50 = 10 $

✔ Answer: 10 — Correct.

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Problem 4:

"An Olympic runner set a record for the 100m dash. The time was ten and sixty-two hundredths seconds. How would you write this as a number?"

"Ten and sixty-two hundredths" = $ 10 + \frac{62}{100} = 10.62 $

✔ Answer: 10.62 seconds — Correct.

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Problem 5:

"The tables at the party are shaped like the hexagon. If you put the tables together, how many would you need for 50 people? Answer: 12 tables"

Assuming standard hexagonal table seating:
- A regular hexagon has 6 sides.
- Typically, when hexagonal tables are joined edge-to-edge, each table can seat 6 people, but shared edges reduce available seats.

But if we assume that each hexagonal table seats 6 people and they are placed separately (not joined), then:

> $ 50 ÷ 6 ≈ 8.33 $ → Need 9 tables

But the answer says 12 tables, which suggests a different configuration.

Wait — maybe the problem assumes only 5 people per table?

Alternatively, perhaps it's about shared sides.

Let’s think differently: In many setups, when hexagonal tables are joined, each additional table adds fewer new seats.

But here's a likely assumption:
- Each individual hexagonal table seats 6 people.
- But if you join tables together, some sides are shared, so fewer total seats.

However, the question says: "If you put the tables together..." — implying they're joined.

But the answer is 12 tables, so let's reverse-engineer:

> 12 tables × ? = 50 people

If 12 tables seat 50 people, then average per table: $ 50 ÷ 12 ≈ 4.17 $ people per table.

That doesn’t make sense unless only 4–5 people per table.

Wait — perhaps the hexagonal tables are arranged in a way that each table seats 5 people?

Alternatively, maybe it's a known puzzle: Hexagonal tables where each table seats 6, but when joined, the outer perimeter determines capacity.

But without a diagram, we must rely on logic.

Another possibility: Each table seats 6 people, and you need to seat 50, so:
> $ 50 ÷ 6 = 8.33 $ → Need 9 tables

But answer says 12, so perhaps there’s a misunderstanding.

Wait — could it be that each table seats 4 or 5 people?

Let’s suppose each table seats 5 people:
> $ 50 ÷ 5 = 10 $ → need 10 tables

Still not 12.

Try: 12 tables → 50 people → average ~4.17 per table.

Perhaps each table seats 4 people?
> $ 50 ÷ 4 = 12.5 $ → need 13 tables

Hmm.

Alternatively, maybe the problem assumes that when tables are joined, only the outer edges are used, and each table contributes 5 seats when joined.

But let’s look for a known pattern.

Actually, in some configurations, when hexagonal tables are joined side-by-side, each table can seat 6 people, but when joined, shared sides lose seats.

But if all tables are separate, then 6 per table → 50 ÷ 6 ≈ 8.3 → 9 tables.

But the answer is 12, so perhaps the problem assumes each table seats only 4 or 5 people.

Wait — maybe it's a trick: Each hexagonal table has 6 sides, and each side seats 1 person, so 6 seats per table.

But when joined, adjacent sides are shared, so fewer total seats.

For example:
- 1 table: 6 seats
- 2 tables joined: 6 + 6 - 2 = 10 seats (since 2 sides are shared)
- 3 tables in a line: 6 + 6 + 6 - 2×2 = 18 - 4 = 14 seats
- Pattern: Each new table adds 4 seats (since 2 sides are shared)

So:
- Table 1: 6
- Table 2: 6 + 4 = 10
- Table 3: 10 + 4 = 14
- Table 4: 18
- Table 5: 22
- Table 6: 26
- Table 7: 30
- Table 8: 34
- Table 9: 38
- Table 10: 42
- Table 11: 46
- Table 12: 50 ✔

Yes! With 12 tables, you get 50 seats.

So the pattern is:
- First table: 6 seats
- Each additional table adds 4 seats (because two sides are shared)

So total seats after $ n $ tables:
> $ S = 6 + 4(n-1) $

Set $ S = 50 $:
> $ 6 + 4(n-1) = 50 $
> $ 4(n-1) = 44 $
> $ n-1 = 11 $
> $ n = 12 $

✔ So 12 tables are needed.

Answer: 12 tables — Correct.

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Problem 6:

"At your birthday party, you had 7 8-slice pizzas. 41 slices were eaten. What fraction of pizza is left?"

Total slices:
> $ 7 × 8 = 56 $

Slices eaten: 41

Slices left:
> $ 56 - 41 = 15 $

Fraction left:
> $ \frac{15}{56} $

But the answer says 5/8 — that’s incorrect.

Wait — let’s check:

Is 5/8 of a pizza left? That would mean:
> $ \frac{5}{8} $ of 56 = $ 56 × \frac{5}{8} = 35 $ slices left

But only 15 slices are left → contradiction.

So 5/8 is wrong.

But wait — maybe the answer is 5/8 of a whole pizza? No — 15/56 ≠ 5/8.

Let’s compute:
> $ \frac{15}{56} $ — simplify?

15 and 56 have no common factors → already simplified.

But the answer says 5/8, which is incorrect.

Wait — perhaps a mistake in the problem or answer?

Let’s double-check:

- 7 pizzas × 8 slices = 56 slices
- 41 eaten → 15 left
- Fraction left = $ \frac{15}{56} $

But the answer says 5/8 — which is about 0.625, while $ \frac{15}{56} ≈ 0.268 $

So clearly not equal.

Unless...

Wait — maybe the answer 5/8 is wrong, or the problem is misread.

Wait — perhaps the answer key is wrong.

But let's see: Is there any way $ \frac{15}{56} = \frac{5}{8} $? No.

$ \frac{5}{8} = \frac{35}{56} $, which would mean 35 slices left, but only 15 were left.

So either:
- The answer is wrong, or
- The problem is misstated.

Wait — maybe 41 slices were eaten, but total slices were less?

No — 7 × 8 = 56.

So 56 - 41 = 15 left.

So fraction left = $ \frac{15}{56} $

But the answer says 5/8 — that’s incorrect.

Wait — could it be asking for fraction of a pizza, not of all pizzas?

No — “what fraction of pizza is left” — probably means fraction of total pizza.

So correct answer should be $ \frac{15}{56} $

But the answer key says 5/8 — which is wrong.

Wait — maybe the problem said "41 slices were eaten", but perhaps only 41 were served?

No — it says had 7 8-slice pizzas, so 56 total.

Wait — unless one pizza was not fully sliced?

No — it says 8-slice pizzas.

So the answer provided (5/8) is incorrect.

But let’s see: Is there a typo?

Suppose only 41 slices were eaten, and total was 56, then 15 left.

But $ \frac{15}{56} $ cannot be simplified to $ \frac{5}{8} $

Wait — unless the answer meant 5/8 of a pizza? But 5/8 of a pizza is 5 slices.

No — 15 slices = $ \frac{15}{8} = 1.875 $ pizzas.

So 1.875 pizzas left, or $ \frac{15}{56} $ of total.

So the correct answer is $ \frac{15}{56} $, not $ \frac{5}{8} $

✘ So the given answer 5/8 is wrong.

But wait — let’s recheck the problem:

> "At your birthday party, you had 7 8-slice pizzas. 41 slices were eaten. What fraction of pizza is left?"

Answer: 5/8

But 41 eaten from 56 → 15 left → $ \frac{15}{56} $

But $ \frac{5}{8} = \frac{35}{56} $, which would mean 35 slices left, so 21 eaten, not 41.

So unless the number of slices eaten is 15, not 41, it’s wrong.

Wait — maybe the answer is 5/8 of a pizza, but that doesn't make sense.

Or maybe the question is: What fraction of a single pizza is left? But that wouldn’t make sense either.

Alternatively, perhaps the answer key is wrong, and the correct answer is 15/56, not 5/8.

But the user says: "Answer— 5/8"

So either:
- There's a typo in the problem or answer
- Or we're missing something

Wait — maybe only 41 slices were served, and 7 pizzas = 56, so 15 not eaten?

No — “41 slices were eaten” implies 41 were consumed.

So 56 - 41 = 15 left.

So fraction left = 15/56

But 5/8 = 35/56 — so unless only 35 slices were eaten, it's wrong.

Wait — perhaps 41 slices were eaten, but only 56 existed, so 15 left.

But 15/56 ≠ 5/8.

So the answer 5/8 is incorrect.

But maybe the problem meant 21 slices were eaten?

Then 56 - 21 = 35 left → $ \frac{35}{56} = \frac{5}{8} $

Ah! So likely there's a typo in the problem or answer.

Either:
- The number eaten is 21, not 41
- Or the answer should be 15/56

But since the user says the answer is 5/8, and the problem says 41 slices eaten, then either the answer is wrong or the problem is misstated.

But based on what's written:

> 7 × 8 = 56
> 56 - 41 = 15
> Fraction left = $ \frac{15}{56} $

So the correct answer is $ \frac{15}{56} $, not $ \frac{5}{8} $

✘ So the given answer is incorrect.

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✔ Final Summary:

| Problem | Given Answer | Correct? | Explanation |
|--------|--------------|---------|------------|
| 1 | 18 | ✔ Yes | GCF=6, LCM=36 → 12×x = 6×36 → x=18 |
| 2 | 72 | ✔ Yes | 5+3+ sum of 1,3,5,...,15 = 8 + 64 = 72 |
| 3 | 10 | ✔ Yes | 20% of 50 = 10 |
| 4 | 10.62 | ✔ Yes | "ten and sixty-two hundredths" = 10.62 |
| 5 | 12 tables | ✔ Yes | Hexagonal tables: 6 + 4(n-1) = 50 → n=12 |
| 6 | 5/8 | ✘ No | Should be $ \frac{15}{56} $, not $ \frac{5}{8} $ |

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🟩 Conclusion:

The only incorrect answer is Problem 6. The correct fraction of pizza left is $ \frac{15}{56} $, not $ \frac{5}{8} $. Likely a typo in the answer key.

Let me know if you'd like help correcting it!